Defining arrays in single line in Javascript
I am trying to declare multiple arrays in a single line. But I am taking this error. (Cannot set property "0" of undefined)
var photos,tags = new Array();
flickrJSON.items.forEach(function(item, index)
photos[index] = item.media.m;
tags[index] = item.tags;
);
Why I take this error? Can somebody explain? and How can I fix it
javascript arrays declaration
asked Nov 15 '18 at 11:28
devneeddevdevneeddev
564
add a comment |
I am trying to declare multiple arrays in a single line. But I am taking this error. (Cannot set property "0" of undefined)
var photos,tags = new Array();
flickrJSON.items.forEach(function(item, index)
photos[index] = item.media.m;
tags[index] = item.tags;
);
Why I take this error? Can somebody explain? and How can I fix it
javascript arrays declaration
asked Nov 15 '18 at 11:28
devneeddevdevneeddev
564
Can you use ES6?
– omri_saadon
Nov 15 '18 at 11:29
Just for the sake of interpreting the error, writing:var photos,tags = new Array();is the same as writingvar photos; var tags = new Array();. Hence the error (photos isundefined).
– briosheje
Nov 15 '18 at 11:34
add a comment |
I am trying to declare multiple arrays in a single line. But I am taking this error. (Cannot set property "0" of undefined)
var photos,tags = new Array();
flickrJSON.items.forEach(function(item, index)
photos[index] = item.media.m;
tags[index] = item.tags;
);
Why I take this error? Can somebody explain? and How can I fix it
javascript arrays declaration
asked Nov 15 '18 at 11:28
devneeddevdevneeddev
564
I am trying to declare multiple arrays in a single line. But I am taking this error. (Cannot set property "0" of undefined)
var photos,tags = new Array();
flickrJSON.items.forEach(function(item, index)
photos[index] = item.media.m;
tags[index] = item.tags;
);
Why I take this error? Can somebody explain? and How can I fix it
javascript arrays declaration
javascript arrays declaration
asked Nov 15 '18 at 11:28
devneeddevdevneeddev
564
asked Nov 15 '18 at 11:28
devneeddevdevneeddev
564
asked Nov 15 '18 at 11:28
devneeddevdevneeddev
564
asked Nov 15 '18 at 11:28
devneeddevdevneeddev
564
asked Nov 15 '18 at 11:28
devneeddevdevneeddev
564
564
Can you use ES6?
– omri_saadon
Nov 15 '18 at 11:29
Just for the sake of interpreting the error, writing:var photos,tags = new Array();is the same as writingvar photos; var tags = new Array();. Hence the error (photos isundefined).
– briosheje
Nov 15 '18 at 11:34
add a comment |
Can you use ES6?
– omri_saadon
Nov 15 '18 at 11:29
Just for the sake of interpreting the error, writing:var photos,tags = new Array();is the same as writingvar photos; var tags = new Array();. Hence the error (photos isundefined).
– briosheje
Nov 15 '18 at 11:34
Can you use ES6?
– omri_saadon
Nov 15 '18 at 11:29
Can you use ES6?
– omri_saadon
Nov 15 '18 at 11:29
Just for the sake of interpreting the error, writing:
var photos,tags = new Array(); is the same as writing var photos; var tags = new Array();. Hence the error (photos is undefined).– briosheje
Nov 15 '18 at 11:34
Just for the sake of interpreting the error, writing:
var photos,tags = new Array(); is the same as writing var photos; var tags = new Array();. Hence the error (photos is undefined).– briosheje
Nov 15 '18 at 11:34
add a comment |
2 Answers
2
active
oldest
votes
You're trying to use two arrays there - an array named photos, and an array named tags, so you need to use new Array (or, preferably, ) twice:
var photos = , tags = ;
In your original code, var photos, will result in photos being undefined, no matter what comes after the comma.
If you wanted to create a lot of arrays at once and didn't want to repeat = each time, you could use Array.from with destructuring to keep code DRY:
const [arr1, arr2, arr3, arr4, arr5] = Array.from( length: 5 , () => );
answered Nov 15 '18 at 11:29
CertainPerformanceCertainPerformance
93.4k165484
it works thank you! I will mark as a true answer.
– devneeddev
Nov 15 '18 at 11:30
add a comment |
Using ES6 you can:
let [photos, tags] = [, ];
answered Nov 15 '18 at 11:30
omri_saadonomri_saadon
7,07841546
add a comment |
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2 Answers
2
active
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2 Answers
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active
oldest
votes
active
oldest
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active
oldest
votes
You're trying to use two arrays there - an array named photos, and an array named tags, so you need to use new Array (or, preferably, ) twice:
var photos = , tags = ;
In your original code, var photos, will result in photos being undefined, no matter what comes after the comma.
If you wanted to create a lot of arrays at once and didn't want to repeat = each time, you could use Array.from with destructuring to keep code DRY:
const [arr1, arr2, arr3, arr4, arr5] = Array.from( length: 5 , () => );
answered Nov 15 '18 at 11:29
CertainPerformanceCertainPerformance
93.4k165484
it works thank you! I will mark as a true answer.
– devneeddev
Nov 15 '18 at 11:30
add a comment |
You're trying to use two arrays there - an array named photos, and an array named tags, so you need to use new Array (or, preferably, ) twice:
var photos = , tags = ;
In your original code, var photos, will result in photos being undefined, no matter what comes after the comma.
If you wanted to create a lot of arrays at once and didn't want to repeat = each time, you could use Array.from with destructuring to keep code DRY:
const [arr1, arr2, arr3, arr4, arr5] = Array.from( length: 5 , () => );
answered Nov 15 '18 at 11:29
CertainPerformanceCertainPerformance
93.4k165484
it works thank you! I will mark as a true answer.
– devneeddev
Nov 15 '18 at 11:30
add a comment |
You're trying to use two arrays there - an array named photos, and an array named tags, so you need to use new Array (or, preferably, ) twice:
var photos = , tags = ;
In your original code, var photos, will result in photos being undefined, no matter what comes after the comma.
If you wanted to create a lot of arrays at once and didn't want to repeat = each time, you could use Array.from with destructuring to keep code DRY:
const [arr1, arr2, arr3, arr4, arr5] = Array.from( length: 5 , () => );
answered Nov 15 '18 at 11:29
CertainPerformanceCertainPerformance
93.4k165484
You're trying to use two arrays there - an array named photos, and an array named tags, so you need to use new Array (or, preferably, ) twice:
var photos = , tags = ;
In your original code, var photos, will result in photos being undefined, no matter what comes after the comma.
If you wanted to create a lot of arrays at once and didn't want to repeat = each time, you could use Array.from with destructuring to keep code DRY:
const [arr1, arr2, arr3, arr4, arr5] = Array.from( length: 5 , () => );
answered Nov 15 '18 at 11:29
CertainPerformanceCertainPerformance
93.4k165484
edited Nov 15 '18 at 11:32
answered Nov 15 '18 at 11:29
CertainPerformanceCertainPerformance
93.4k165484
answered Nov 15 '18 at 11:29
CertainPerformanceCertainPerformance
93.4k165484
answered Nov 15 '18 at 11:29
CertainPerformanceCertainPerformance
93.4k165484
93.4k165484
it works thank you! I will mark as a true answer.
– devneeddev
Nov 15 '18 at 11:30
add a comment |
it works thank you! I will mark as a true answer.
– devneeddev
Nov 15 '18 at 11:30
it works thank you! I will mark as a true answer.
– devneeddev
Nov 15 '18 at 11:30
it works thank you! I will mark as a true answer.
– devneeddev
Nov 15 '18 at 11:30
add a comment |
Using ES6 you can:
let [photos, tags] = [, ];
answered Nov 15 '18 at 11:30
omri_saadonomri_saadon
7,07841546
add a comment |
Using ES6 you can:
let [photos, tags] = [, ];
answered Nov 15 '18 at 11:30
omri_saadonomri_saadon
7,07841546
add a comment |
Using ES6 you can:
let [photos, tags] = [, ];
answered Nov 15 '18 at 11:30
omri_saadonomri_saadon
7,07841546
Using ES6 you can:
let [photos, tags] = [, ];
answered Nov 15 '18 at 11:30
omri_saadonomri_saadon
7,07841546
answered Nov 15 '18 at 11:30
omri_saadonomri_saadon
7,07841546
answered Nov 15 '18 at 11:30
omri_saadonomri_saadon
7,07841546
answered Nov 15 '18 at 11:30
omri_saadonomri_saadon
7,07841546
7,07841546
add a comment |
add a comment |
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Can you use ES6?
– omri_saadon
Nov 15 '18 at 11:29
Just for the sake of interpreting the error, writing:
var photos,tags = new Array();is the same as writingvar photos; var tags = new Array();. Hence the error (photos isundefined).– briosheje
Nov 15 '18 at 11:34